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Dmitriy789 [7]
2 years ago
14

Write 16.24 as a fraction in simplest form

Mathematics
1 answer:
cricket20 [7]2 years ago
4 0

Answer:

2/3

Step-by-step explanation:

16/24 = 14/ 21 = 12/ 18 10/ 15 8/12 6/9 4/6 = 2/3

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nataly862011 [7]
Square roots are the result of multiplying the number by itself for example take 5x5 = 25 that’s a result of a number when multiplying by itself.
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GET IT RIGHT OR GET BANNED LOL!!​
frutty [35]

Answer:

The last one

Step-by-step explanation:

The last one shows x being multiplied by 0.28, which is the product

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the high school debate team is developing a logo to represent their club. A scale drawing of the logo design is presented below,
Fantom [35]

Answer:

The enlargement will be 25 times and the enlargement area will be $2250 \text{ inches}^2$.

Step-by-step explanation:

It is given that each grid unit is equal to 3 inches. SO we have to use this scale.

The total height of the scale drawing is 21 inch and the enlargement has a height given as 105 inches. Therefore it has scale factor of 5. It means that each dimension is enlarged by 5 times the dimension in the scale drawing. So the enlargement of the logo will be 25 times and the enlargement area will be 2250 square inches.

7 0
2 years ago
Evaluate.<br> 21 + } + (-)) -<br> Enter your answer as a mixed number in simplest form in the box.
andrew-mc [135]

Evaluate: 2(1/4) + 2/5 ÷ (-1/2) - 1/4

⇛9/4 + 2/5 ÷ (-1/2) - 1/4

Take the LCM of the 4 and 5 is 20.

Again, 2 and 4 is 4.

⇛[(54+8)/20] ÷ [(-2 - 1)/4]

⇛62/20 ÷ (-3/4)

⇛62/20 × 4/-3

⇛62/{5*(-3)}

⇛62/-15.

3 0
2 years ago
Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)
Triss [41]

g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}

The surface element is

\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

and the integral is

\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)

###

To compute the last integral, you can integrate by parts:

u=r\implies\mathrm du=\mathrm dr

\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}

\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr

For this integral, consider a substitution of

r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds

\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds

\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds

=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds

and the result above follows.

4 0
3 years ago
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